# A truck pulls boxes up an incline plane. The truck can exert a maximum force of #3,200 N#. If the plane's incline is #(5 pi )/8 # and the coefficient of friction is #3/7 #, what is the maximum mass that can be pulled up at one time?

We're asked to find the maximum mass a truck can pull up at one time, with some given information.

The maximum mass the truck can pull is the largest so that the net force is not down the incline (then the boxes would not be going up, intuitively). Thus, at this condition, the boxes are in equilibrium (no net forces acting, pulls upward with constant velocity).

Therefore,

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The maximum mass that can be pulled up the incline can be calculated using the formula:

[ m = \frac{F_{\text{max}} - F_{\text{friction}}}{g \sin(\theta) + \mu_k g \cos(\theta)} ]

where:

- ( F_{\text{max}} ) is the maximum force the truck can exert (3,200 N),
- ( F_{\text{friction}} ) is the force of friction (( \mu_k \cdot F_{\text{normal}} )),
- ( \theta ) is the angle of the incline (5π/8),
- ( \mu_k ) is the coefficient of friction (3/7),
- ( g ) is the acceleration due to gravity (9.8 m/s²).

Plugging in the given values:

[ m = \frac{3200 - \frac{3}{7} \cdot mg \cos\left(\frac{5\pi}{8}\right)}{g \sin\left(\frac{5\pi}{8}\right) + \frac{3}{7} \cdot g \cos\left(\frac{5\pi}{8}\right)} ]

[ m = \frac{3200 - \frac{3}{7} \cdot mg \cos\left(\frac{5\pi}{8}\right)}{g \sin\left(\frac{5\pi}{8}\right) + \frac{3}{7} \cdot g \cos\left(\frac{5\pi}{8}\right)} ]

[ m = \frac{3200 - \frac{3}{7} \cdot (9.8) \cdot m \cdot \cos\left(\frac{5\pi}{8}\right)}{(9.8) \cdot \sin\left(\frac{5\pi}{8}\right) + \frac{3}{7} \cdot (9.8) \cdot \cos\left(\frac{5\pi}{8}\right)} ]

[ m = \frac{3200 - \frac{29.4}{7} \cdot m \cdot \cos\left(\frac{5\pi}{8}\right)}{9.8 \cdot \sin\left(\frac{5\pi}{8}\right) + \frac{29.4}{7} \cdot \cos\left(\frac{5\pi}{8}\right)} ]

Solving for ( m ):

[ m = \frac{3200}{9.8 \cdot \sin\left(\frac{5\pi}{8}\right) + \frac{29.4}{7} \cdot \cos\left(\frac{5\pi}{8}\right) + \frac{29.4}{7}} ]

[ m \approx \frac{3200}{9.8 \cdot 0.9239 + \frac{29.4}{7} \cdot (-0.3827) + \frac{29.4}{7}} ]

[ m \approx \frac{3200}{9.049 + (-3.059) + 4.2} ]

[ m \approx \frac{3200}{10.19} ]

[ m \approx 314.37 , \text{kg} ]

Therefore, the maximum mass that can be pulled up at one time is approximately 314.37 kg.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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